# Deriving charge for Noether current in free complex scalar field QFT

by Azelketh
Tags: charge, complex, current, deriving, field, free, noether, scalar
 P: 36 1. The problem statement, all variables and given/known data Hi I a attempting to derive the expression for the conserved Noether charge for a free complex scalar field. The question I have to complete is: " show, by using the mode expansions for the free complex scalar field, that the conserved Noether charge (corresponding to complex phase rotations) is given by $$Q= \int \frac{\partial^3 p}{(2\pi)^3} ( a_p^* a_p - b_p^* b_p)$$ Where $$a_p, a_p^*$$ and $$b_p, b_p^*$$ are the creation and annihilation operators for 2 kinds of particles respectively" 2. Relevant equations The Lagrangian for a free complex scalar field is; $$L= \frac{1}{2} \partial_μ \psi \partial^μ \psi- \frac{1}{2}m^2\psi^2$$ I have recognized that there is a symmetry of$$\psi → e^{ i\alpha } \psi$$ which leads to a Noether current of $$j_μ = i \partial _μ \psi^* \psi - i \partial _μ \psi \psi^*$$ $$Q= \int \partial ^3 x j_0 = i \int \partial ^3 x ( \partial _0 \psi^* \psi - \partial _0 \psi \psi^* )$$ $$Q=i \int \partial ^3 x ( \psi \dot{\psi^*} - \psi^* \dot{\psi} )$$ where $$\dot{\psi} = \pi ^*$$ $$\dot{\psi^*} = \pi$$ so $$Q=i \int \partial ^3 x ( \psi \pi - \psi^* \pi ^* )$$ And the mode expansions; $$\psi = \int \frac{\partial^3 P}{(2\pi)^3} \frac{1}{\sqrt{2 ω_p}}( a_p e^{ip.x } + b_p^* e^{-ip.x})$$ $$\psi^* = \int \frac{\partial^3 P}{(2\pi)^3} \frac{1}{\sqrt{2 ω_p}}( a_p^* e^{-ip.x} + b_p e^{ip.x })$$ $$\pi = i\int \frac{\partial^3 P}{(2\pi)^3} \sqrt{ \frac{ω_p}{2}}( a_p^* e^{-ip.x } - b_p e^{ip.x})$$ $$\pi^* = -i\int \frac{\partial^3 P}{(2\pi)^3} \sqrt{\frac{ω_p}{2}}( a_p e^{ip.x } - b_p^* e^{-ip.x})$$ 3. The attempt at a solution I have then plugged these into the Q equation; $$Q=i \int \partial ^3 x [ \int \frac{\partial^3 P}{(2\pi)^3} \frac{1}{\sqrt{2 ω_p}}( a_p e^{ip.x } + b_p^* e^{-ip.x}) i\int \frac{\partial^3 P}{(2\pi)^3} \sqrt{ \frac{ω_p}{2}}( a_p^* e^{-ip.x } - b_p e^{ip.x}) - \int \frac{\partial^3 P}{(2\pi)^3} \frac{1}{\sqrt{2 ω_p}}( a_p^* e^{-ip.x} + b_p e^{ip.x }) (-i)\int \frac{\partial^3 P}{(2\pi)^3} \sqrt{\frac{ω_p}{2}}( a_p e^{ip.x } - b_p^* e^{-ip.x}) ]$$ so factoring out the i's and 2 pi's from the integrals; $$Q=\frac{i^2}{(2\pi)^6} \int \partial ^3 x [ \int \partial^3 P \frac{1}{\sqrt{2 ω_p}}( a_p e^{ip.x } + b_p^* e^{-ip.x}) \int \partial^3 P \sqrt{ \frac{ω_p}{2}}( a_p^* e^{-ip.x } - b_p e^{ip.x}) + \int \partial^3 P \frac{1}{\sqrt{2 ω_p}}( a_p^* e^{-ip.x} + b_p e^{ip.x }) \int \partial^3 P \sqrt{\frac{ω_p}{2}}( a_p e^{ip.x } - b_p^* e^{-ip.x}) ]$$ It's at this point I'm utterly stuck on how to procede, I did think that could cancel the sqrt ω terms but they depend on p: $$ω_p = \sqrt{P^2 + m^2}$$ so as they are inside integrals dependent on P they can't be cancelled. I am at an utter loss how to proceed from here though. If anyone can offer any pointers or assistance is would be greatly appreciated. EDIT: made a sign error in exponential's for $$\psi$$ and $$\psi^*$$
 HW Helper Thanks ∞ PF Gold P: 4,476 You can try changing the order of integration and doing the x integration before the p integrations.
 P: 36 Ok, I see how you can change the order of the integration, and changing one of the P's to r to make it clearer to me, to get : $$Q=\frac{-1}{(2\pi)^6} \int \partial ^3 x [ \int \partial^3 P \frac{1}{\sqrt{2 ω_p}}( a_p e^{ip.x } + b_p^* e^{-ip.x}) \int \partial^3 r \sqrt{ \frac{ω_r}{2}}( a_r^* e^{-ir.x } - b_r e^{ir.x}) + \int \partial^3 P \frac{1}{\sqrt{2 ω_p}}( a_p^* e^{-ip.x} + b_p e^{ip.x }) \int \partial^3 r \sqrt{\frac{ω_r}{2}}( a_r e^{ir.x } - b_r^* e^{-ir.x}) ]$$ $$Q=\frac{-1}{(2\pi)^6} \int \partial ^3 x [ \int \partial^3 P \sqrt{\frac{ω_r}{ ω_p}} ( a_p e^{ip.x } + b_p^* e^{-ip.x}) \int \partial^3 r ( a_r^* e^{-ir.x } - b_r e^{ir.x}) + \int \partial^3 P \sqrt{\frac{ω_r}{ ω_p}}( a_p^* e^{-ip.x} + b_p e^{ip.x }) \int \partial^3 r ( a_r e^{ir.x } - b_r^* e^{-ir.x}) ]$$ $$Q=\frac{-1}{(2\pi)^6} ( \int \partial^3 P \int \partial^3 r \int \partial ^3 x \sqrt{\frac{ω_r}{ ω_p}}( a_p e^{ip.x } + b_p^* e^{-ip.x} ) ( a_r^* e^{-ir.x } - b_r e^{ir.x}) + \int \partial^3 P \int \partial^3 r \int \partial ^3 x \sqrt{\frac{ω_r}{ ω_p}}( a_p^* e^{-ip.x} + b_p e^{ip.x })( a_r e^{ir.x } - b_r^* e^{-ir.x}) )$$ $$Q=\frac{-1}{(2\pi)^6} ( \int \partial^3 P \int \partial^3 r \int \partial ^3 x \sqrt{\frac{ω_r}{ ω_p}} ( a_p a_r^* e^{i( p - r ).x} -a_p b_r e^{i( p + r).x} + b_p^* a_r^* e^{-i( p + r ).x} -b_p^* b_r e^{i( r - p ).x} ) + \int \partial^3 P \int \partial^3 r \int \partial ^3 x \sqrt{\frac{ω_r}{ ω_p}} ( a_p^* a_r e^{i( r - p ).x} -a_p^* b_r^* e^{-i( P + r ).x} + b_p a_re^{i( p + r).x} - b_p b_r^* e^{i( p - r ).x} )$$ I'm not sure how to go about integrating the x components; I've never integrated anything over $$\partial^3 x$$ and am struggling to find anything that can help explain it. would; $$\int \partial ^3 x a_p^* a_r e^{ i(p +r).x} = \frac{1}{ (i(p+r))^3}a_p^* a_r e^{ i(p +r).x}$$ or as its a dot product in the exponential, $$\int \partial ^3 x a_p^* a_r e^{ i(p +r).x} = \int \partial ^3 x a_p^* a_r e^{ i( (p1 +r1)x1 + (p2+r2)x2 +(p3+r3)x3 )}$$ $$= \frac{1}{i^3(p1+r1)(p2+r2)(p3+r3) }a_p^* a_r e^{ i(p +r).x}$$ But i'm sort of expecting the x integral to give some kind of delta function in r so that the r integral is easy leaving just the p integral ie the required result. EDIT: maybe the commutation relations for the ap and bp operators will give delta functions, now to get the x integral out of the way first! EDIT2: corrected sign error in exponential's.
HW Helper
Thanks ∞
PF Gold
P: 4,476

## Deriving charge for Noether current in free complex scalar field QFT

The delta function will come in through a well known "identity" that you can find here http://dlmf.nist.gov/1.17 (see equation 1.17.12).

Before going further, make sure you have the correct signs in your exponential functions for ##\pi## and ##\pi^*## (that you used to construct ##Q##).
 P: 36 ah, I think its $$\psi$$ and $$\psi^*$$ that have the incorrect signs. Will edit through all above to correct this. Bloody sign errors, when will they end!
 P: 36 ahh finally solved it, using $$\int \partial^3 x e^{i(p+r).x} = (2\pi)^3 \delta^3(p+r)$$ and working through. Thanks for the assitance TSny!

 Related Discussions Advanced Physics Homework 13 Advanced Physics Homework 0 Classical Physics 3 Quantum Physics 2 Advanced Physics Homework 6